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- Question 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed vo enters in between the horizontal plates of a capacitor. The electric field strength between the plates of length L and distance d, is E. The vertical deviation of the electron at the moment of exit from the field region is measured to be Y. Derive the expression giving the electron's charge to mass ratio, i.e. e/m to be 2v,Y/CEL). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10* Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10* km/s. (Be careful to use coherent units.)arrow_forwardThe figure shows the distribution of nuclear charges (positive charges) in a KBr molecule. Find the magnitude of the electric field (at 10¹0 N/C) at the center of mass of the molecule, knowing that dBr=9.3.10-11 dk-1.89.10-1⁰ m. e=1,602.10-¹⁹℃) a) 506,76 Br +35e b) 516,76 dBr (k=9.10⁹ c.m. dk Fig.123 c) 526,76 K +19e d) 536,76 m, N•m²/C² e) 546,76 "arrow_forwardE = E1 E = 0 E= E1 E= E1 The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure above. On the right side of the cube, the electric field has magnitude Ej = 426 V/m, and the angle between the electric field and the surface of the cube is 0 = 15 degrees. On the bottom of the cube, the electric field has the same magnitude Ej, and the angle between the electric field and the surface of the cube is also 0 = 15 degrees. On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has magnitude Ej and is parallel to the face; on the other half of the front and back faces, the electric field is zero. One edge of the cube is 41 cm long. Part 1 What is the net electric flux on this cubical surface? Net electric flux = i V•m Save for Later Attempts: 0 of 4 used Submit Answerarrow_forward
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- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial distancer from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = Edr = - Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C= Q I Vabarrow_forwardCalculate the magnitude of strength of an electric field at the distance of 12.0 nm from the nucleus of the Uranium-235 atom?arrow_forwardThe plane z = 0 marks the boundary between free space (z 0 side) with a relative permittivity of e, =35. The electric field intensity next to the interface in free space is Ē = –10£ + 25ŷ + 92 V/m. Determine the electric field intensity on the other side of the interface.arrow_forward
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